Primitive of Reciprocal of x cubed plus a cubed

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Theorem

$\ds \int \frac {\d x} {x^3 + a^3} = \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$


Proof

\(\ds \int \frac {\d x} {x^3 + a^3}\) \(=\) \(\ds \int \paren {\frac 1 {3 a^2 \paren {x + a} } - \frac {x - 2 a} {3 a^2 \paren {x^2 - a x + a^2} } } \rd x\) Partial Fraction Expansion
\(\ds \) \(=\) \(\ds \int \paren {\frac 1 {3 a^2 \paren {x + a} } - \frac {2 x - 4 a} {6 a^2 \paren {x^2 - a x + a^2} } } \rd x\) multiplying top and bottom by $2$
\(\ds \) \(=\) \(\ds \int \paren {\frac 1 {3 a^2 \paren {x + a} } - \paren {\frac {2 x - a} {6 a^2 \paren {x^2 - a x + a^2} } } - \paren {\frac {3 a} {6 a^2 \paren {x^2 - a x + a^2} } } } \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 {3 a^2} \int \frac {\d x} {x + a} - \frac 1 {6 a^2} \int \frac {\paren {2 x - a} \rd x} {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {3 a^2} \ln \size {x + a} - \frac 1 {6 a^2} \int \frac {\paren {2 x - a} \rd x} {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}\) Primitive of $\dfrac 1 {a x + b}$
\(\ds \) \(=\) \(\ds \frac 1 {3 a^2} \ln \size {x + a} - \frac 1 {6 a^2} \ln \size {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}\) Primitive of Function under its Derivative
\(\ds \) \(=\) \(\ds \frac 1 {6 a^2} \ln \size {\paren {x + a}^2} - \frac 1 {6 a^2} \ln \size {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}\) Logarithm of Power
\(\ds \) \(=\) \(\ds \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}\) Difference of Logarithms
\(\ds \) \(=\) \(\ds \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {2 a} \paren {\frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} } }\) Primitive of $\dfrac 1 {x^2 - a x + a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}\) simplifying

$\blacksquare$


Sources

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